Prove that $|a_{n+1}-a_{n}| \leq \frac{1}{2}|a_{n}-a_{n-1}|$ implies $\{a_n\}$ bounded and convergent.

Question

I am trying to prove that every sequence for which

$|a_{n+1}-a_{n}| \leq \frac{1}{2}|a_{n}-a_{n-1}|$

is true, is bounded and convergent. I tried defining $b_n = a_n$ and $c_n = a_{n-1}$ and then use Stolz, but it didn't work.

How could I prove this?

Actually you can prove that for any $0<C<1$ instead of $\frac{1}{2}$ it holds that sequence is a Cauchy sequence. — arberavdullahu, Dec 24 '16 at 14:09
The sequence is Cauchy. It will be convergent for sure if the sequence is defined in an order complete space like $\Bbb R$. — Masacroso, Dec 24 '16 at 14:21

score 3 · Answer 1 · answered Dec 24 '16 at 14:12

Define $c:=|a_1-a_0|\ge 0$. Then by induction $|a_{n+1}-a_{n}|\le \frac{c}{2^n}$, therefore $(x_n)_n$ is a Cauchy sequence indeed $$ |x_n-x_m|\le |x_n-x_{n-1}|+|x_{n-1}-x_{n-2}|+\cdots+|x_{m+1}-x_m| \le \frac{c}{2^{m-1}} $$ for all $n>m$. Then it is enough to choose $m$ sufficiently large so that $|x_n-x_m|<\varepsilon$.

Clement C. · Accepted Answer · 2016-12-24T14:30:04.697

3

For the boundedness:

You can write, using the triangle inequality and your assumption,$^{(\dagger)}$ $$ \lvert a_n - a_0 \rvert = \left\lvert\sum_{k=0}^{n-1} (a_{k+1}-a_k)\right\rvert \leq \sum_{k=0}^{n-1} \left\lvert a_{k+1}-a_k\right\rvert \leq \sum_{k=0}^{n-1} \frac{1}{2^{k}}\left\lvert a_{1}-a_0\right\rvert \leq \sum_{k=0}^{\infty} \frac{1}{2^{k}}\left\lvert a_{1}-a_0\right\rvert = 2\left\lvert a_{1}-a_0\right\rvert $$ so we get $\lvert a_n\rvert \leq \lvert a_n - a_0 \rvert + \lvert a_0\rvert \leq 2\lvert a_1 - a_0 \rvert + \lvert a_0\rvert$ for all $n$.

$(\dagger)$ Indeed, by a straightforward induction we get $\left\lvert a_{n+1}-a_n\right\rvert \leq \frac{1}{2^{n}}\left\lvert a_{1}-a_0\right\rvert$ for all $n\geq 0$.

(for the convergence, the same "trick" but starting the sum at some $m$ (i.e., with $a_m$ instead of $a_0$) will enable you to show the sequence is Cauchy.)

edited Dec 24 '16 at 14:30

answered Dec 24 '16 at 14:15

Clement C.

67,323

I'm trying your solution right now – ItamarG3 Dec 24 '16 at 14:17
thank you! I never thought about that 3rd step! – ItamarG3 Dec 24 '16 at 14:23
Glad this helped. – Clement C. Dec 24 '16 at 14:24
1

Actually $\sum_{k=0}^{\infty} \frac{1}{2^{k}}=2$. – Janik Dec 24 '16 at 14:28
Darn, indeed. Messing up numbers. (Thanks!) – Clement C. Dec 24 '16 at 14:29
@ClementC. I'm not sure I did the 3rd step correctly for the convergence. It's supposed to be $\leq \sum_{k=m}^{n-1}\frac{1}{2^k}|a_{m+1}-a_m|$, right? – ItamarG3 Dec 24 '16 at 14:39
(From my phone) yes, this looks believable. Now, take the part that doesn't depend on k out of the sum, and sum what remains - the 1/2^k - from k=m to (upper bound) infinity. – Clement C. Dec 24 '16 at 14:42
@ClementC. thanks! – ItamarG3 Dec 24 '16 at 14:43

score 1 · Answer 3 · answered Dec 24 '16 at 14:14

1

Hint: Try $$ |a_n|=|a_1+(a_2-a_1)++\cdots+(a_{n-1}-a_{n-2})+(a_n-a_{n-1})|\le|a_1|+|a_2-a_1|+\cdots+|a_{n-1}-a_{n-2}|+|a_n-a_{n-1}|$$

answered Dec 24 '16 at 14:14

xpaul

44,000

Prove that $|a_{n+1}-a_{n}| \leq \frac{1}{2}|a_{n}-a_{n-1}|$ implies $\{a_n\}$ bounded and convergent.

3 Answers3

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